3.7.56 \(\int \frac {-312+72 x+(108 x-24 x^2) \log (5)+(-9 x^2+2 x^3) \log ^2(5)+e^x (6+6 x-x^2 \log (5))}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx\)

Optimal. Leaf size=25 \[ x+(-10+x) x+\frac {2+e^x}{\frac {6}{x}-\log (5)} \]

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Rubi [B]  time = 0.51, antiderivative size = 156, normalized size of antiderivative = 6.24, number of steps used = 15, number of rules used = 8, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {27, 6742, 43, 77, 2199, 2194, 2177, 2178} \begin {gather*} x^2-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}-\frac {24 x}{\log (5)}+\frac {6 e^x}{\log (5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}-\frac {e^x}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-312 + 72*x + (108*x - 24*x^2)*Log[5] + (-9*x^2 + 2*x^3)*Log[5]^2 + E^x*(6 + 6*x - x^2*Log[5]))/(36 - 12*
x*Log[5] + x^2*Log[5]^2),x]

[Out]

x^2 - E^x/Log[5] - (24*x)/Log[5] + 432/(Log[5]^2*(6 - x*Log[5])) - 312/(Log[5]*(6 - x*Log[5])) + (6*E^x)/(Log[
5]*(6 - x*Log[5])) - (108*(4 - Log[125]))/(Log[5]^2*(6 - x*Log[5])) + (3*x*(8 - Log[125]))/Log[5] + (72*Log[6
- x*Log[5]])/Log[5]^2 + (108*(2 - Log[5])*Log[6 - x*Log[5]])/Log[5]^2 - (36*(8 - Log[125])*Log[6 - x*Log[5]])/
Log[5]^2

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{(-6+x \log (5))^2} \, dx\\ &=\int \left (-\frac {312}{(-6+x \log (5))^2}+\frac {72 x}{(-6+x \log (5))^2}-\frac {12 x (-9+2 x) \log (5)}{(-6+x \log (5))^2}+\frac {x^2 (-9+2 x) \log ^2(5)}{(-6+x \log (5))^2}-\frac {e^x \left (-6-6 x+x^2 \log (5)\right )}{(-6+x \log (5))^2}\right ) \, dx\\ &=-\frac {312}{\log (5) (6-x \log (5))}+72 \int \frac {x}{(-6+x \log (5))^2} \, dx-(12 \log (5)) \int \frac {x (-9+2 x)}{(-6+x \log (5))^2} \, dx+\log ^2(5) \int \frac {x^2 (-9+2 x)}{(-6+x \log (5))^2} \, dx-\int \frac {e^x \left (-6-6 x+x^2 \log (5)\right )}{(-6+x \log (5))^2} \, dx\\ &=-\frac {312}{\log (5) (6-x \log (5))}+72 \int \left (\frac {6}{\log (5) (-6+x \log (5))^2}+\frac {1}{\log (5) (-6+x \log (5))}\right ) \, dx-(12 \log (5)) \int \left (\frac {2}{\log ^2(5)}+\frac {18 (4-\log (125))}{\log ^2(5) (6-x \log (5))^2}+\frac {3 (-8+\log (125))}{\log ^2(5) (6-x \log (5))}\right ) \, dx+\log ^2(5) \int \left (\frac {2 x}{\log ^2(5)}-\frac {108 (-2+\log (5))}{\log ^3(5) (-6+x \log (5))}+\frac {3 (8-\log (125))}{\log ^3(5)}-\frac {108 (-4+\log (125))}{\log ^3(5) (-6+x \log (5))^2}\right ) \, dx-\int \left (\frac {e^x}{\log (5)}-\frac {6 e^x}{(-6+x \log (5))^2}+\frac {6 e^x}{\log (5) (-6+x \log (5))}\right ) \, dx\\ &=x^2-\frac {24 x}{\log (5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}+6 \int \frac {e^x}{(-6+x \log (5))^2} \, dx-\frac {\int e^x \, dx}{\log (5)}-\frac {6 \int \frac {e^x}{-6+x \log (5)} \, dx}{\log (5)}\\ &=x^2-\frac {6 e^{\frac {6}{\log (5)}} \text {Ei}\left (-\frac {6-x \log (5)}{\log (5)}\right )}{\log ^2(5)}-\frac {e^x}{\log (5)}-\frac {24 x}{\log (5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}+\frac {6 e^x}{\log (5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}+\frac {6 \int \frac {e^x}{-6+x \log (5)} \, dx}{\log (5)}\\ &=x^2-\frac {e^x}{\log (5)}-\frac {24 x}{\log (5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}+\frac {6 e^x}{\log (5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 47, normalized size = 1.88 \begin {gather*} \frac {-12-e^x x \log (5)+x^3 \log ^2(5)+18 x \log (125)-3 x^2 \log (5) (2+\log (125))}{\log (5) (-6+x \log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-312 + 72*x + (108*x - 24*x^2)*Log[5] + (-9*x^2 + 2*x^3)*Log[5]^2 + E^x*(6 + 6*x - x^2*Log[5]))/(36
- 12*x*Log[5] + x^2*Log[5]^2),x]

[Out]

(-12 - E^x*x*Log[5] + x^3*Log[5]^2 + 18*x*Log[125] - 3*x^2*Log[5]*(2 + Log[125]))/(Log[5]*(-6 + x*Log[5]))

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fricas [B]  time = 0.60, size = 49, normalized size = 1.96 \begin {gather*} -\frac {x e^{x} \log \relax (5) - {\left (x^{3} - 9 \, x^{2}\right )} \log \relax (5)^{2} + 6 \, {\left (x^{2} - 9 \, x\right )} \log \relax (5) + 12}{x \log \relax (5)^{2} - 6 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*log(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*log(5)^2+(-24*x^2+108*x)*log(5)+72*x-312)/(x^2*log(5)^2-12
*x*log(5)+36),x, algorithm="fricas")

[Out]

-(x*e^x*log(5) - (x^3 - 9*x^2)*log(5)^2 + 6*(x^2 - 9*x)*log(5) + 12)/(x*log(5)^2 - 6*log(5))

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giac [B]  time = 0.39, size = 52, normalized size = 2.08 \begin {gather*} \frac {x^{3} \log \relax (5)^{2} - 9 \, x^{2} \log \relax (5)^{2} - 6 \, x^{2} \log \relax (5) - x e^{x} \log \relax (5) + 54 \, x \log \relax (5) - 12}{x \log \relax (5)^{2} - 6 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*log(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*log(5)^2+(-24*x^2+108*x)*log(5)+72*x-312)/(x^2*log(5)^2-12
*x*log(5)+36),x, algorithm="giac")

[Out]

(x^3*log(5)^2 - 9*x^2*log(5)^2 - 6*x^2*log(5) - x*e^x*log(5) + 54*x*log(5) - 12)/(x*log(5)^2 - 6*log(5))

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maple [A]  time = 0.23, size = 35, normalized size = 1.40

method	result	size
norman	\(\frac {x^{3} \ln \relax (5)+52 x +\left (-9 \ln \relax (5)-6\right ) x^{2}-{\mathrm e}^{x} x}{x \ln \relax (5)-6}\)	\(35\)
risch	\(x^{2}-9 x -\frac {12}{\ln \relax (5) \left (x \ln \relax (5)-6\right )}-\frac {x \,{\mathrm e}^{x}}{x \ln \relax (5)-6}\)	\(35\)
default	\(-\frac {12}{\ln \relax (5) \left (x \ln \relax (5)-6\right )}-\frac {6 \,{\mathrm e}^{x}}{\ln \relax (5)^{2} \left (x -\frac {6}{\ln \relax (5)}\right )}-9 x +x^{2}-\frac {{\mathrm e}^{x}}{\ln \relax (5)}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2*ln(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*ln(5)^2+(-24*x^2+108*x)*ln(5)+72*x-312)/(x^2*ln(5)^2-12*x*ln(5)+3
6),x,method=_RETURNVERBOSE)

[Out]

(x^3*ln(5)+52*x+(-9*ln(5)-6)*x^2-exp(x)*x)/(x*ln(5)-6)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -{\left (\frac {432}{x \log \relax (5)^{5} - 6 \, \log \relax (5)^{4}} - \frac {x^{2} \log \relax (5) + 24 \, x}{\log \relax (5)^{3}} - \frac {216 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{4}}\right )} \log \relax (5)^{2} + 9 \, {\left (\frac {36}{x \log \relax (5)^{4} - 6 \, \log \relax (5)^{3}} - \frac {x}{\log \relax (5)^{2}} - \frac {12 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{3}}\right )} \log \relax (5)^{2} + 24 \, {\left (\frac {36}{x \log \relax (5)^{4} - 6 \, \log \relax (5)^{3}} - \frac {x}{\log \relax (5)^{2}} - \frac {12 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{3}}\right )} \log \relax (5) - 108 \, {\left (\frac {6}{x \log \relax (5)^{3} - 6 \, \log \relax (5)^{2}} - \frac {\log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{2}}\right )} \log \relax (5) - \frac {x e^{x}}{x \log \relax (5) - 6} - \frac {6 \, e^{\frac {6}{\log \relax (5)}} E_{2}\left (-\frac {x \log \relax (5) - 6}{\log \relax (5)}\right )}{{\left (x \log \relax (5) - 6\right )} \log \relax (5)} - \frac {432}{x \log \relax (5)^{3} - 6 \, \log \relax (5)^{2}} + \frac {312}{x \log \relax (5)^{2} - 6 \, \log \relax (5)} + \frac {72 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{2}} - 6 \, \int \frac {e^{x}}{x^{2} \log \relax (5)^{2} - 12 \, x \log \relax (5) + 36}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*log(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*log(5)^2+(-24*x^2+108*x)*log(5)+72*x-312)/(x^2*log(5)^2-12
*x*log(5)+36),x, algorithm="maxima")

[Out]

-(432/(x*log(5)^5 - 6*log(5)^4) - (x^2*log(5) + 24*x)/log(5)^3 - 216*log(x*log(5) - 6)/log(5)^4)*log(5)^2 + 9*
(36/(x*log(5)^4 - 6*log(5)^3) - x/log(5)^2 - 12*log(x*log(5) - 6)/log(5)^3)*log(5)^2 + 24*(36/(x*log(5)^4 - 6*
log(5)^3) - x/log(5)^2 - 12*log(x*log(5) - 6)/log(5)^3)*log(5) - 108*(6/(x*log(5)^3 - 6*log(5)^2) - log(x*log(
5) - 6)/log(5)^2)*log(5) - x*e^x/(x*log(5) - 6) - 6*e^(6/log(5))*exp_integral_e(2, -(x*log(5) - 6)/log(5))/((x
*log(5) - 6)*log(5)) - 432/(x*log(5)^3 - 6*log(5)^2) + 312/(x*log(5)^2 - 6*log(5)) + 72*log(x*log(5) - 6)/log(
5)^2 - 6*integrate(e^x/(x^2*log(5)^2 - 12*x*log(5) + 36), x)

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mupad [B]  time = 0.22, size = 30, normalized size = 1.20 \begin {gather*} -\frac {x\,\left (6\,x+{\mathrm {e}}^x+9\,x\,\ln \relax (5)-x^2\,\ln \relax (5)-52\right )}{x\,\ln \relax (5)-6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((72*x + exp(x)*(6*x - x^2*log(5) + 6) + log(5)*(108*x - 24*x^2) - log(5)^2*(9*x^2 - 2*x^3) - 312)/(x^2*log
(5)^2 - 12*x*log(5) + 36),x)

[Out]

-(x*(6*x + exp(x) + 9*x*log(5) - x^2*log(5) - 52))/(x*log(5) - 6)

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sympy [A]  time = 0.24, size = 31, normalized size = 1.24 \begin {gather*} x^{2} - 9 x - \frac {x e^{x}}{x \log {\relax (5 )} - 6} - \frac {12}{x \log {\relax (5 )}^{2} - 6 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2*ln(5)+6*x+6)*exp(x)+(2*x**3-9*x**2)*ln(5)**2+(-24*x**2+108*x)*ln(5)+72*x-312)/(x**2*ln(5)**2
-12*x*ln(5)+36),x)

[Out]

x**2 - 9*x - x*exp(x)/(x*log(5) - 6) - 12/(x*log(5)**2 - 6*log(5))

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